Question

What is this compound empirical formula?

54.53% C, 9.15% H, 36.32% O by mass

also if its molar mass is 132 amu, what is its molecular formula

Answers

(0.5453)*(132/12) = ~ 6

(0.0915)*(132) = ~ 12

(0.3632)*(132/16)= ~ 3

C2H4O is the empirical while

C6H12O3 is the molecular

#1

80%

#2

C6H12O3

Probably a chain C6H11O2OH.

#3

in 1 mole of the compound we have 54.53% C or 54.53 % of 132 = 72 g of c orr 72/ 12 or 6 moles of C

mass of H = 9.15 % of 132 = 12g or 12 moles of H

mass of O = 36.32 % of 132 = 48g of O in 1 mole of the compound = 48/16 moles = 3

molecular formula = C6H12O3 ..

#4